Residual conformal symmetry in GSW

[Rene Meyer]

Hello,

I hope you don't mind this elementary question, but I got stuck with
it this evening:

GSW state in their equ. (2.1.44) that a combined reparametrisation and
a Weyl scaling obeying

d^aA^b + d^bA^a = Gamma n^ab

don't change the choosen gauge h^ab=n^ab. Now I tried to proof this.
The metric varies according to

delta n^ab = Gamma n^ab

and this should be zero. As n^ab is definitly not zero, Gamma = 0.
Then d^aA^b = - d^bA^a, which automatically implies eq. (2.1.19) to
vanish too. But then, in the coordinates sigma^+ and sigma^-,

d/d^sigma^+ A^+ = 0 = d/d^sigma^- A^-

Thus the statement that A^+=A^+(sigma^+) and similarily for A^- is
wrong, though I know is it right from other sources. So what is wrong
with my line of reasoning?

Hope someone can find some time and help me.

René.

--
René Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China

[Lubos Motl]

On Mon, 29 Mar 2004, Rene Meyer wrote:

> GSW state in their equ. (2.1.44) that a combined reparametrisation and
> a Weyl scaling obeying
>
> d^aA^b + d^bA^a = Gamma n^ab
>
> don't change the choosen gauge h^ab=n^ab. Now I tried to proof this...

Dear Rene,

I am afraid that there are too many errors in your reasoning, and I won't
comment on your argument because already your starting point is
incorrect, and therefore all other steps are incorrect, too.

The first fact that you don't seem to appreciate (or know) is that under a
small diffeomorphism given by an infinitesimal vector field A^a, the
metric changes by

delta h^{ab} = d^a A^b + d^b A^a ($#) where "d" is the covariant derivative (one can also lower the indices if one changes the sign). You seem to say that it (or everything) changes either by zero or by a multiple of itself, which suggests that you are not aware of ($#). It is not hard to prove ($#), for example by Taylor-expanding the general formula for the transformation of a tensor. h^{ab} (x) = h^{a'b'} (x(x')) . M^a_a' . M^b_b. wbere M^a_a' is the matrix of derivatives of x^a with respect to x'^a' - in this case you want to choose x'^a = x^a + A^a (where A^a is infinitesimal). Let me just motivate ($#). It has the right structure of indices; it is
dimensionless as the metric should be (A^a cancels the derivative) and it
uses covariant derivatives. And it is ab-symmetric. And if "A^2" depends
linearly on "x^2", for example, you know that you are "stretching" the
coordinate "x^2", and therefore "h^{22}" will change by a constant, which
is exactly what ($#) says. The tensor field d^a A^b + d^b A^a is a general tensor field, but if you choose the vector field A^a appropriately, it may equal a multiple of the metric that you started with (at each point of your manifold). In this case you can "undo" the change of the metric by making a Weyl transformation. Under a Weyl transformation, the metric changes by a multiple of itself, i.e. delta h^{ab} = epsilon . h^{ab} () The total variation of the metric is the sum of the ($#) piece and the
($$$) piece, and for a special choice of A^a and epsilon, this total
variation can vanish.

Best wishes
Lubos
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